3.969 \(\int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx\)

Optimal. Leaf size=107 \[ -\frac {\left (4 a c+3 b^2\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{16 c^{5/2}}-\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c} \]

[Out]

-1/16*(4*a*c+3*b^2)*arctan(1/2*(-2*c*x^2+b)/c^(1/2)/(-c*x^4+b*x^2+a)^(1/2))/c^(5/2)-3/8*b*(-c*x^4+b*x^2+a)^(1/
2)/c^2-1/4*x^2*(-c*x^4+b*x^2+a)^(1/2)/c

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Rubi [A]  time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1114, 742, 640, 621, 204} \[ -\frac {\left (4 a c+3 b^2\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{16 c^{5/2}}-\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5/Sqrt[a + b*x^2 - c*x^4],x]

[Out]

(-3*b*Sqrt[a + b*x^2 - c*x^4])/(8*c^2) - (x^2*Sqrt[a + b*x^2 - c*x^4])/(4*c) - ((3*b^2 + 4*a*c)*ArcTan[(b - 2*
c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 - c*x^4])])/(16*c^(5/2))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )\\ &=-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c}-\frac {\operatorname {Subst}\left (\int \frac {-a-\frac {3 b x}{2}}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c}+\frac {\left (3 b^2+4 a c\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )}{16 c^2}\\ &=-\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c}+\frac {\left (3 b^2+4 a c\right ) \operatorname {Subst}\left (\int \frac {1}{-4 c-x^2} \, dx,x,\frac {b-2 c x^2}{\sqrt {a+b x^2-c x^4}}\right )}{8 c^2}\\ &=-\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c}-\frac {\left (3 b^2+4 a c\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{16 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 89, normalized size = 0.83 \[ -\frac {\left (4 a c+3 b^2\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{16 c^{5/2}}-\frac {\left (3 b+2 c x^2\right ) \sqrt {a+b x^2-c x^4}}{8 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/Sqrt[a + b*x^2 - c*x^4],x]

[Out]

-1/8*((3*b + 2*c*x^2)*Sqrt[a + b*x^2 - c*x^4])/c^2 - ((3*b^2 + 4*a*c)*ArcTan[(b - 2*c*x^2)/(2*Sqrt[c]*Sqrt[a +
b*x^2 - c*x^4])])/(16*c^(5/2))

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fricas [A]  time = 0.81, size = 211, normalized size = 1.97 \[ \left [-\frac {{\left (3 \, b^{2} + 4 \, a c\right )} \sqrt {-c} \log \left (8 \, c^{2} x^{4} - 8 \, b c x^{2} + b^{2} - 4 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {-c} - 4 \, a c\right ) + 4 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} + 3 \, b c\right )}}{32 \, c^{3}}, -\frac {{\left (3 \, b^{2} + 4 \, a c\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {c}}{2 \, {\left (c^{2} x^{4} - b c x^{2} - a c\right )}}\right ) + 2 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} + 3 \, b c\right )}}{16 \, c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/32*((3*b^2 + 4*a*c)*sqrt(-c)*log(8*c^2*x^4 - 8*b*c*x^2 + b^2 - 4*sqrt(-c*x^4 + b*x^2 + a)*(2*c*x^2 - b)*sq
rt(-c) - 4*a*c) + 4*sqrt(-c*x^4 + b*x^2 + a)*(2*c^2*x^2 + 3*b*c))/c^3, -1/16*((3*b^2 + 4*a*c)*sqrt(c)*arctan(1
/2*sqrt(-c*x^4 + b*x^2 + a)*(2*c*x^2 - b)*sqrt(c)/(c^2*x^4 - b*c*x^2 - a*c)) + 2*sqrt(-c*x^4 + b*x^2 + a)*(2*c
^2*x^2 + 3*b*c))/c^3]

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giac [A]  time = 0.21, size = 91, normalized size = 0.85 \[ -\frac {1}{8} \, \sqrt {-c x^{4} + b x^{2} + a} {\left (\frac {2 \, x^{2}}{c} + \frac {3 \, b}{c^{2}}\right )} - \frac {{\left (3 \, b^{2} + 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {-c} x^{2} - \sqrt {-c x^{4} + b x^{2} + a}\right )} \sqrt {-c} + b \right |}\right )}{16 \, \sqrt {-c} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-1/8*sqrt(-c*x^4 + b*x^2 + a)*(2*x^2/c + 3*b/c^2) - 1/16*(3*b^2 + 4*a*c)*log(abs(2*(sqrt(-c)*x^2 - sqrt(-c*x^4
+ b*x^2 + a))*sqrt(-c) + b))/(sqrt(-c)*c^2)

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maple [A]  time = 0.02, size = 120, normalized size = 1.12 \[ -\frac {\sqrt {-c \,x^{4}+b \,x^{2}+a}\, x^{2}}{4 c}+\frac {a \arctan \left (\frac {\left (x^{2}-\frac {b}{2 c}\right ) \sqrt {c}}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{4 c^{\frac {3}{2}}}+\frac {3 b^{2} \arctan \left (\frac {\left (x^{2}-\frac {b}{2 c}\right ) \sqrt {c}}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{16 c^{\frac {5}{2}}}-\frac {3 \sqrt {-c \,x^{4}+b \,x^{2}+a}\, b}{8 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(-c*x^4+b*x^2+a)^(1/2),x)

[Out]

-1/4*x^2*(-c*x^4+b*x^2+a)^(1/2)/c-3/8*b*(-c*x^4+b*x^2+a)^(1/2)/c^2+3/16*b^2/c^(5/2)*arctan((x^2-1/2*b/c)/(-c*x
^4+b*x^2+a)^(1/2)*c^(1/2))+1/4*a/c^(3/2)*arctan((x^2-1/2*b/c)/(-c*x^4+b*x^2+a)^(1/2)*c^(1/2))

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maxima [A]  time = 2.42, size = 105, normalized size = 0.98 \[ -\frac {\sqrt {-c x^{4} + b x^{2} + a} x^{2}}{4 \, c} - \frac {3 \, b^{2} \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{16 \, c^{\frac {5}{2}}} - \frac {a \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{4 \, c^{\frac {3}{2}}} - \frac {3 \, \sqrt {-c x^{4} + b x^{2} + a} b}{8 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(-c*x^4 + b*x^2 + a)*x^2/c - 3/16*b^2*arcsin(-(2*c*x^2 - b)/sqrt(b^2 + 4*a*c))/c^(5/2) - 1/4*a*arcsin
(-(2*c*x^2 - b)/sqrt(b^2 + 4*a*c))/c^(3/2) - 3/8*sqrt(-c*x^4 + b*x^2 + a)*b/c^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^5}{\sqrt {-c\,x^4+b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^2 - c*x^4)^(1/2),x)

[Out]

int(x^5/(a + b*x^2 - c*x^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{\sqrt {a + b x^{2} - c x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(-c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral(x**5/sqrt(a + b*x**2 - c*x**4), x)

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